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Math Help - log equation

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    1

    log equation

    I am stuck on the following problems. I don't know how to 'plug in,' or substitute, the correct way--please show me how.

    1) The yield V is in millions of cubic feet and x is the number of years. What is the time needed to yield 2.1 million cubic feet?

    V(x)=6.7(e^-48.1/x)

    2) With buying power B and t years, when will B(t) be half of what it is today?

    B(t)=(0.92)^t

    For this one my teacher said that since the original buying power was 1 I had to change it to .5. She said I should put the .5 in front of the (0.92) but I still don't know what to do from there.

    Also for this one logarithm, I think I've done it right, can you verify?

    1) logx^2 = -10
    x^2 = 10^-10
    x = 100000
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  2. #2
    Banned
    Joined
    Nov 2007
    Posts
    54

    Flip your answer on the log problem.

    "I think I've done it right, can you verify?
    1) logx^2 = -10
    x^2 = 10^-10
    x = 100000"

    log x^2 = -10
    10^-10 = some number between 0 and 1.
    In fact, x^2 = 1/10,000,000,000.
    So x = sqrt(10^-10) = 1/100,000.
    You need to invert your answer.
    If 0 < x^2 < 1 then x < 1.
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